Consider 4NH3(g)+5O2(g) <>4NO(g)+6H2O(g). Delta H=-904.4 kJ. How does each of the following changes affect the yield of NO at equilibrium? Increase, decrease or no changea) increase [NH3]b) increase [H2O]c) decrese [O2]d) decrease the volume of the container in which the reaction occurse) add a catalystf) increase temperature

a) increase; b) decrease; c) decrease; d) decrease; e) same; b) decrease

Explanation:

4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g); ΔH = -904.4 kJ  

The enthalpy of reaction is negative, so the reaction is exothermic, and we can consider the heat as a "product".

We can rewrite the equation as

4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g) + 904.4 kJ  

According to Le Châtelier's Principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

a) Increase [NH₃]

If we increase the concentration of ammonia, the position of equilibrium (POE) will move to the right to get rid of the added ammonia. The yield of NO will increase.

b) Increase [H₂O]

If we add increase the concentration of water vapour, the POE will move to the left to get rid of the added water. The yield of NO will decrease.

c) Decrease [O₂]

If we add decrease the concentration of oxygen, the POE will move to the left to get to replace the removed oxygen. The yield of NO will decrease.

d) Decrease the volume of the container

We have nine moles of reactants and 10 mol of products.

If we decrease volume of the container, the POE will move to the left (the direction of fewer moles) to reduce the increased pressure. The yield of NO will decrease.

e) Add a catalyst

Adding a catalyst will speed up the approach to equilibrium, but it has no effect on the POE.  The POE will remain unchanged. The yield of NO will stay the same.

f) Increase temperature

If we increase the temperature, the POE will move to the left to get rid of the added heat. The yield of NO will decrease.

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: